Engines that transmit an offset piston's power to a straight power shaft have been attempted since at least 1921, e.g. U.S. Pat. No. 1,365,666 but have not had practical success though they inherently offer high torque and high fuel efficiency. Their weakness lies in using many energy absorbing moving parts and combustion chambers to convert the piston's reciprocating rectilinear motion to the power shaft's unidirectional rotary motion which has made them inefficient and impractical, e.g. U.S. Pat. Nos. 2,239,663; and 5,673,665. For this reason, the simple, exhaust polluting, inefficient but reliable crankshaft engine survives as the search for a better power source continues.
Enormous funds and research have been poured into fuel cells, electric vehicles and crank engine hybrids for years to replace the ubiquitous crank engine.
The crank engine is very inefficient because the two angles at both ends of the connecting rod of length L and the crank angle α (FIG. 10) combine to slow the piston's speed, which traps the very rapidly expanding combustion gases in a small chamber. The gases build up very high heat and pressure at and near tdc. Here, nearly all the force from the pressure is vectored against the crankshaft's bearings instead of rotating it. Parts inertia is combined with extra fuel on each power stroke to overcome the angles' resistance. The result is excess exhaust pollution and waste heat. The waste heat is lost and the pollutants are partly scrubbed from the exhaust when it is too late.
A higher rpm increases efficiency but that has reached its limit and it is not good enough. The pollution and the waste heat must be reduced in the combustion chamber by converting them to mechanical motion with a more complete burn. To do that, all the rod and crank angles must be zero during the entire power stroke but that is impossible in a crank engine. The following mathematics explain why:
FIG. 10 is a schematic that represents a crank engine. FV1, FV2, FV3 are force vectors that come from burn pressure driving the piston 38. FV1 is along a radial of the crankshaft axis C. Only FV3, being tangent to the crank circle d, rotates the shaft where FV3=FV1(Cos θ)(Cos Φ). The crank engine's efficiency is zero at tdc when angle θ=0 but angle Φ=90°, making FV3=FV1(1)(0)=0. When FV2 is tangent to circle d, Cos Φ=1.0 and Tan θ=r/L and θ=Tan−1r/L from which Cos θ is found. The efficiency at that point is FV3/FV1=Cos θ. The importance of angle θ=Tan−1r/L is explained below.
The ratio of the displacement M along the crank circle d to the piston's displacement a at any chosen crank angle θ is easily found from FIG. 10. r is the crank arm length and α is in degrees:r=b+a a=r(1−Cos α)M=παr/180M/a=πα/[180(1−Cos α)]For instance, when α=10°, M/a=11.49:1. At this point, the rod's slow crank end must go 11.49 times as far as the piston. The slower the crank's rotation, the longer the gases are trapped in a small chamber and the lower the engine's efficiency. It is known that this is where the confined hot, pressurized gases create most of the pollution and waste heat. The crank's angular efficiency:Cos θ=FV2/FV1Cos Φ=FV3/FV2FV2=FV1(Cos θ)FV2=FV3/Cos ΦFV3=FV1(Cos θ)(Cos Φ)FV3/FV1=(Cos θ)(Cos Φ) Crank engine's angular efficiency. It caps thermal efficiency.
FIG. 10 is also the basis for the following indented equations that lead to the Cos θ and Cos Φ equations in terms of crank angle α, length L and crank arm r:180−β=γγ+θ+Φ=180β=90−α Note the rt. triangle (α+β+90)180−(90−α)=γ or 90+α=γ(90+α)+θ+Φ=180α+θ+Φ=90n=r Sin αSin θ=(r/L)Sin αθ=Sin−1[(r/L)Sin α]Cos θ=Cos{Sin−1[(r/L)Sin α]}α+Sin−1[(r/L)Sin α]+Φ=90Φ=90−{α+Sin−1[(r/L)Sin α]}Cos Φ=Cos(90−{α+Sin−1[(r/L)Sin α]})The equations Cos θ, Cos Φ are easily solved with a hand calculator.
The importance of angle θ=Tan−1r/L now follows. That is when FV2 is tangent to the circle d at the arm r which makes angle Φ=0.0 and Cos Φ=1.0. The angular efficiency is Cos θ=Cos(Tan−1r/L). Extend L relative to r so that angle θ goes to 0.0. Then
                                1        ⁢          Lim              θ        →        0.0              ⁢                  ⁢    Cos    ⁢                  ⁢    θ    =      1.0    .  That makes the angular efficiency FV3/FV1=Cos θ)(Cos Φ)=(1)(1)=100% because there is no angular resistance since the angles θ, Φ disappear. The variable angle α disappears. The crank arm r disappears. The variable length torque arm n (FIG. 10) which requires torque buildup is replaced by the fixed length torque arm r′ (FIG. 11) which gives instant peak torque. 1 This is the foundation for calculus.
Unlike the crank, FV1 in this invention (FIG. 11) is always directed to rotating the output shaft 8 rather than directed against the shaft's bearings. FV1 is transmitted with both angles θ, Φ=0.0 through the entire power stroke. The M/a=1:1 through the entire stroke. The circumference d′ replaces the crank circle d in FIG. 10. Shaft 8 receives shear force alone from the fixed length torque arm r′.